#pragma GCC optimize("O3") #include using namespace std; using ll=long long; using P=pair; template using V=vector; #define fi first #define se second #define all(v) (v).begin(),(v).end() const ll inf=(1e18); const ll mod=998244353; //const ll mod=1000000007; ll GCD(ll a,ll b) {return b ? GCD(b,a%b):a;} ll LCM(ll c,ll d){return c/GCD(c,d)*d;} struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout< bool chmax(T &a, const T &b) { if (a bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; } struct mint{ using ull=unsigned long long int; ull v; mint(ll vv=0){s(vv%mod+mod);} mint& s(ull vv){ v=vv>=1; } return res; } mint inv()const{return pow(mod-2);} //拡張ユークリッドの互除法 /* mint inv()const{ int x,y; int g=extgcd(v,mod,x,y); assert(g==1); if(x<0)x+=mod; return mint(x); }*/ friend ostream& operator<<(ostream&os,const mint&val){ return os<(const mint&val)const{return v>val.v;} }; int main(){ int t; cin>>t; ll n,k; while(t--){ cin>>n>>k; mint ans=0; if(n<=k){ ans=mint(2).pow(k*(n-1))*mint(n)*(mint(2).pow(k)-mint(1)); }else{ ans=mint(n+1).pow(k-2)*mint(n)*mint(n*2+1); } cout<